Compact Set in Metric Space


by wayneckm
Tags: compact, metric, space
wayneckm
wayneckm is offline
#1
Oct24-10, 09:36 PM
P: 68
Hello all,


Here is my question while reading a proof.

For a compact set [tex] K [/tex] in a separable metrizable spce [tex] (E,\rho) [/tex] and a continuous function [tex] t \mapsto f(t) [/tex], if we define

[tex] D_{K} = \inf \{ t \geq 0 \; : \; f(t) \in K \}[/tex]

then, [tex] D_{K} \leq t [/tex] if and only if [tex] \inf\{ \rho(f(q),K) : q \in \mathbb{Q} \cap [0,t] \}[/tex] = 0

May someone shed some light on this? I do not understand it. Thanks very much.


Wayne
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Landau
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#2
Oct25-10, 06:28 AM
Sci Advisor
P: 905
It's not really clear what the domain and codomain of your function f are.
wayneckm
wayneckm is offline
#3
Oct25-10, 07:00 AM
P: 68
Domain of [tex] f [/tex] is [tex] \mathbb{R}^{+} [/tex]
Codomain of [tex] f [/tex] is [tex] \mathbb{R} [/tex]

Landau
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#4
Oct25-10, 07:05 AM
Sci Advisor
P: 905

Compact Set in Metric Space


Huh? Then I don't understand [itex]f(t)\in K[/itex]...
wayneckm
wayneckm is offline
#5
Oct25-10, 04:33 PM
P: 68
Oops...sorry, i misunderstood the term codomain. So codomain here should be [tex] E [/tex] as stated.
Office_Shredder
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#6
Oct25-10, 05:02 PM
Mentor
P: 4,499
If [tex] D_k \leq t[/tex] then [tex] f(D_k)\in K[/tex]. We can approximate [tex] D_k[/tex] with rational numbers, and because [tex] D_k \in [0,t][/tex] we can approximate [tex] D_k[/tex] with rational numbers in [tex] \mathbb{Q}\cap [0,t][/tex] If [tex] q_r[/tex] is such a sequence converging to [tex] D_k[/tex], the distance between [tex] f(q_r)[/tex] and [tex] f(D_k)[/tex] goes to zero, which means the distance between [tex]f(q_r)[/tex] and [tex]K[/tex] must go to zero. So the infimum of the distance between [tex] f(q)[/tex] and [tex]K[/tex] for [tex] q\in \mathbb{Q}\cap [0,t][/tex] must be zero because we just found a sequence for which the distance is arbitrarily small.

This is basically the direction [tex] D_k\leq t[/tex] implies the infimum is zero.


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